trying to compute the integral $\int_\gamma \frac{e^{z}}{z(z-3)} dz $, where $\gamma:[0,2\pi]\to\mathbb{C}, \gamma(\theta)=2e^{i\theta} $ but not sure where to begin. I know, from Cauchy's formula, that $ f(b) = \frac{1}{2{\pi}i} \int_\gamma \frac{f(a)}{a-b}da $, so I attempted to proceed by letting $ f(a) = e^{a} $ and realizing the denominator consists of product terms $ z $ and $z-3$, both of the form $a-b$, with $a=z$ and $b=0$ and $b=3$ in each term. However, I'm not sure how to deal with the two product terms, nor am I sure in my $ f(a) = e^{a} $ substitution. Any help is greatly appreciated!

Observe that $|3|>2$, hence $g(z)=\frac{e^z}{z-3}$ is analytic in $|z|<3$. So you can rewrite you integral in the form familiar from Cauchy's theorem: (applying the theorem to $g(z)$)
$$\int_\gamma\frac{g(z)}{z-0}dz=2\pi ig(0)=-\frac{1}{3}$$